Gravity Waves in a Flowing Fluid

Consider a gravity wave traveling through a fluid that is flowing horizontally at the uniform velocity ${\bf V}=V\,{\bf e}_x$ . Let us write

$${\bf v}({\bf r},t) = {\bf V} + {\bf v}_1({\bf r},t),$$ (11.84)

$$p({\bf r},t) =p_0 - \rho\,g\,z + p_1({\bf r},t),$$ (11.85)

where ${\bf v}_1$ and $p_1$ are the small velocity and pressure perturbations, respectively, due to the wave. To first order in small quantities, the fluid equations of motion, (11.1) and (11.2), reduce to

$$\nabla \cdot{\bf v}_1 = 0,$$ (11.86)

$$\left(\frac{\partial}{\partial t} + {\bf V}\cdot\nabla\right){\bf v}_1 = -\frac{\nabla p_1}{\rho},$$ (11.87)

respectively. We can also define the displacement, $\xi$ $({\bf r},t)$ , of a fluid particle due to the passage of the wave, as seen in a frame co-moving with the fluid, as

$$\left(\frac{\partial}{\partial t} + {\bf V}\cdot\nabla\right) \mbox{\boldmath$\xi$} = {\bf v}_1.$$ (11.88)

The curl of Equation (11.87) implies that $\nabla\times{\bf v}_1={\bf0}$ . Hence, we can write ${\bf v}_1=-\nabla\phi$ , and Equation (11.87) yields

$$\left(\frac{\partial}{\partial t} + {\bf V}\cdot\nabla\right)\phi = \frac{p_1}{\rho}.$$ (11.89)

Finally, Equation (11.86) gives

$$\nabla^{\,2}\phi = 0.$$ (11.90)

The most general traveling wave solution to Equation (11.90), with wave vector ${\bf k}=k\,{\bf e}_x$ , and angular frequency $\omega$ , is

$$\phi(x,z,t)= \left[A\,\cosh(k\,z)+B\,\sinh(k\,z)\right]\cos(\omega\,t-k\,x).$$ (11.91)

It follows from Equation (11.89) that

$$p_1(x,z,t) = \rho\,k\,(V-c)\left[A\,\cosh(k\,z) + B\,\sinh(k\,z)\right]\sin(\omega\,t-k\,x),$$ (11.92)

and from Equation (11.88) that

$$\xi_z(x,z,t,) = (V-c)^{-1}\left[A\,\sinh(k\,z)+ B\,\cosh(k\,z)\right]\sin(\omega\,t-k\,x).$$ (11.93)

Here, $c=\omega/k$ is the phase velocity of the wave.